@Yoav Kallus; You are right... second failed attempt on my side. And in fact your idea is conceptually much simpler. For a smaller (coordinate-wise) example, your $k_3$ can be changed to any vector with odd sum of coordinates and different from $(\pm 1,0)$ or $(0,\pm 1)$. (E.g, $k_3=(3,0)$ or $k_3=(2,1)$ work).

No. My "fundamental unit" is the Minkowski sum of the five vectors I state. All its edges (as vectors) are parallel to one of the five defining vectors. (Less important, there was a mistake in my comment. I intended $P$ to be $[0,2]^3$ so that it is indeed a lattice polytope since I assumed that is one requirement you pose. So its lattice points are the eight vertices $\{0,2\}^3$ of the cube plus the six centers of the facets of the cube).

I still have problems with your definitions, but maybe the following is a counter-example to your question: Let $P$ be the cube $[-1,1]^3$, with respect to the lattice $L=\{(x,y,z)\in \Z^3 : x+y+z \in 2\Z\}$. Take as "fundamental unit" the (Minkowski sum of) the vectors $(2,0,0)$, $(0,2,0)$, (0,0,2)$, $(3,1,0)$ and $(0,3,1)$.

You should clearly define what you mean by "fundamental unit". Do you mean "set of non-zero shortest vectors in the lattice"?

Any four of the five triangulations of the pentagon give an example of a "non-realizable family". More generally, any set of triangulations whose union is the complete graph but which does not consist of the whole set of triangulations is not realizable...

I would say no "combinatorial" description that distinguishes regular from non-regular triangulations of $\Delta^n\times \Delta^k$ is known. For example, my paper cited by Jo O'Rourke above contains a nice description and a counting algorithm for triangulations of $\Delta^2\times \Delta^k$, but none of them distinguish the regular from the non-regular. In terms of asymptotics, in the same paper I show that the number of regular triangulations of $\Delta^n\times \Delta^k$ is less than $\left(\frac{e}{2}kn\right)^{n(n-1)(k-1)}$.

There is a conjecture stronger than the original one and weaker than the (false and) more general one that might be true: Conjecture: For every flag simplicial polytope, the maximum size of a set of pairwise intersecting facets is achieved by the facets containing some common vertex.

3. A more constructive proof of the same Lemma is as follows. As before, suppose $P$ is a simplex. Any point $p$ in $dP$ can be written as a (perhaps not integral) combination $\lambda_1 v_1 + \cdots + \lambda_{n+1} v_{n+1}$ of the vertices of $P$, with $\sum_i \lambda_i=d$ and all $\lambda_i\ge 0$. Since $d\ge n+1$, some $\lambda_i$ is $\ge 1$. That means that $p - v_i$ is in $(d-1)P$. Iterate until you get a point at height $\le n$.